Abstract

It is well known that traveling salesman problem has the exp(Ω(n)) time complexity in most cases. The exponential base will be reduced with the bounded degree graphs. The number of Hamiltonian cycles is approximately computed as e×(Davg/2)n for a n-vertex graph of average degree Davg, where e is the base of natural logarithm. It is much smaller than the total number of Hamiltonian cycles (n−1)!/2 in a complete undirected graph. An approximate method to compute a sparse graph is introduced to decrease the search space of the optimal solution. Firstly, a set of optimal k-vertex paths are computed with a weighted graph. Secondly, a frequency graph is computed with the set of optimal k-vertex paths. The frequency graph maintains the topological structure of the weighted graph whereas the numbers on the edges represent the frequencies of edges enumerated form the set of optimal k-vertex paths. Thirdly, a frequency threshold is deduced to remove the edges with frequencies below it and the sparse graph is generated. The bounded degree of the sparse graph is much smaller than n−1 and the complexity of traveling salesman problem is reduced. The approximate method is verified with tens of Euclidean TSP instances.

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